You connect a 250-$\Omega$ resistor,a 1.20-mH inductor,and a 1.80-$\mu$F capacitor in series across a 60.0-Hz,120-V (peak)source.The rms current in your circuit is approximately
A)56.9 mA
B)80.5 mA
C)0.480 mA
D)37.8 mA
E)165 mA

Question

Quizplus · Accepted Answer

The rms current can be calculated using the formula: I_rms = V_rms / Z, where Z is the impedance of the circuit. First, calculate the reactance of the inductor (X_L = 2πfL) and the capacitor (X_C = 1/(2πfC)). Then, find the impedance Z = √(R^2 + (X_L - X_C)^2). Finally, use V_rms = V_peak/√2 to find I_rms. The calculations lead to approximately 56.9 mA.

You Connect a 250- Ω\OmegaΩ Resistor,a 1 μ\muμ F Capacitor in Series Across a 60

You Connect a 250- $\Omega$ Resistor,a 1 $\mu$ F Capacitor in Series Across a 60